Problem Description
裸的区间增减的题目,由于数据有点大,得用long long。输入n,m 有n个元素,m次询问。C a b c 将区间a-b的数全部加c,Q a b,求区间a-b的和
代码:
#include<cstdio>
using namespace std;
struct node
{
long long num, lazy;
};
node tree[];
#define MID int mid = (l + r) / 2
//预处理
#define lson root<<1
#define rson root<<1|1
node merge_(node x, node y)
//回潮求和
{
node t;
t.lazy = ;
//因为这里没有初始化,wrong了 好多遍
t.num = x.num + y.num;
return t;
}
void build(int root, int l, int r)
//初始化线段树
{
tree[root].lazy = ;
//初始化0
if(l == r)
{
scanf("%lld", &tree[root].num);
//输入
return ;
}
MID;
build(lson, l, mid);
//左子树递归
build(rson, mid + , r);
//右子树递归
tree[root] = merge_(tree[lson], tree[rson]);
}
void pushdown(int root, int l, int r)
//向下更新
{
if(tree[root].lazy) {//如果lazy有值的话
tree[lson].lazy += tree[root].lazy;
//左孩子lazy 值更新
tree[rson].lazy += tree[root].lazy;
//右孩子lazy 值更新
MID;
tree[lson].num += (mid - l + ) * tree[root].lazy;//更新和
tree[rson].num += (r - (mid + ) + ) * tree[root].lazy;
//更新和
tree[root].lazy = ;
//变为0
}
}
void updata(int root, int l, int r, int ul, int ur, long long v)
//区间ul-ur增加v
{
if(ul <= l && r <= ur)
//在区间内
{
tree[root].lazy += v;
tree[root].num += (r - l + ) * v;
return ;
}
pushdown(root, l, r);
//向下更新lazy
MID;
if(mid >= ul) updata(lson, l, mid, ul, ur, v);
//递归
if(mid < ur) updata(rson, mid + , r, ul, ur, v);
//递归
tree[root] = merge_(tree[lson], tree[rson]);
//归并
}
long long query(int root, int l, int r, int ul, int ur)
//求ul-ur的和
{
if(ul <= l && r <= ur)
//在区间内返回值
{
return tree[root].num;
}
pushdown(root, l, r);
MID;
long long red = ;
if(mid >= ul) red = query(lson, l, mid, ul, ur);
//mid 在 左孩子区间
if(mid < ur) ryou query(rson, mid + , r, ul, ur);
//mid 在 右孩子区间
return red;
}
int main()
{
int n, m, L, R;
long long v;
char c;
while(~scanf("%d %d", &n, &m))
{
build(, , n);
while(m--)
{
scanf("%*c%c", &c);
if(c == 'Q')
{
scanf("%d %d", &L, &R);
printf("%lld\n", query(, , n, L, R));
}
else
{
scanf("%d %d %lld", &L, &R, &v);
updata(, , n, L, R, v);
}
}
}
return ;
}